Answer:
Graph B express the magnetic relationship of magnetic flux and electronic flow
A 200-lb man carries a 10-lb can of paint up a helical staircase that encircles a silo with radius 30 ft. If the silo is 60 ft high and the man makes exactly two complete revolutions, how much work is done by the man against gravity in climbing to the top
Answer:
17.07 kJ
Explanation:
The work done against gravity by the man W equals the potential energy change of the man and can of paint, ΔU
W = ΔU = mgΔy where m = mass of man and can of paint = 200 lb + 10 lb = 210 lb = 210 × 1 kg/2.205 lb, g = acceleration due to gravity = 9.8 m/s² and Δy = height of silo = 60 ft = 60 × 1m/3.28 ft
Since W = mgΔy, we substitute the values of the variables into the equation.
So,
W = mgΔy
W = 210 lb × 1 kg/2.205 lb × 9.8 m/s² × 60 ft × 1m/3.28 ft
W = 123480/7.2324 J
W = 17073.2 J
W = 17.0732 kJ
W ≅ 17.07 kJ
A gymnast falls from a height onto a trampoline. For a moment, both the gymnast’s kinetic energy and gravitational potential energy are zero. How is the gymnast’s mechanical energy stored for that moment? Question 12 options: rest energy chemical energy elastic energy thermal energy
Answer:
elastic energy
Explanation:
When a gymnast falls on a trampoline from a height, after coming in contact with the trampoline, both the gymnast and the trampoline start to move down due to the elastic property of the trampoline.
During this stretching of the trampoline there comes a maximum point up to which the trampoline is stretched. At this point, both the kinetic energy and the gravitational potential energy of the gymnast are zero due to zero speed and zero height, respectively.
The only energy stored in the gymnast's body at this point is the elastic potential energy due to stretching of the trampoline. Hence,the correct option is:
elastic energy
A baseball of mass 0.145 kg is thrown at a speed of 40.0 m/s. The batter strikes the ball with a force of 15,000 N; the bat and ball are in contact for 0.500 ms. The force is exactly opposite to the original direction of the ball. Determine the final speed of the ball.
The final speed of the ball is 91.72 m/s.
Given the following data:
Mass of baseball = 0.145 kgInitial speed = 40.0 m/sForce = 15,000 NewtonTime = 0.500 milliseconds (ms) to seconds = 0.0005 seconds.To find the final speed of the ball, we would use the following formula:
[tex]F = \frac{M(V - U)}{t}[/tex]
Where:
F is the force applied. u is the initial speed. v is the final speed. t is the time measured in seconds.Substituting the parameters into the formula, we have;
[tex]15000 = \frac{0.145(V \;- \;40)}{0.0005}\\\\15000(0.0005) = 0.145(V \;- \;40)\\\\7.5 = 0.145V - 5.8\\\\0.145V = 7.5 + 5.8\\\\0.145V = 13.3\\\\V = \frac{13.3}{0.145}[/tex]
Final speed, V = 91.72 m/s
Therefore, the final speed of the ball is 91.72 m/s.
Read more here: https://brainly.com/question/24029674
1.a machine gun fires a ball with an initial velocity of 600m/s with an elevation of 30° with respect to the ground neglecting air resistance calculate:
a.the maximum height that can be reached?
b.the time of flight of the bullet?
c.the maximum horizontal displacement of the ired bullet?
Answer:
See explanation
Explanation:
a) maximum height of a projectile = u sin^2θ/2g
H= 600 × (sin 30)^2/2 × 10
H= 7.5 m
b) Time of flight
t= 2u sinθ/g
t= 2 × 600 sin 30/10
t= 60 seconds
Range
R= u^2sin2θ/g
R= (600)^2 × sin2(30)/10
R= 31.2 m
Explore the Prisms screen to see how your understanding of refraction applies when light travels through a medium like glass. Give specific examples and images from the simulation to explain how your understanding applies
Explanation:
https://tse2.mm.bing.net/th?id=OGC.b52c959ac810db1177599a161631c917&pid=Api&rurl=https%3a%2f%2fupload.wikimedia.org%2fwikipedia%2fcommons%2fthumb%2ff%2ff5%2fLight_dispersion_conceptual_waves.gif%2f266px-Light_dispersion_conceptual_waves.gif&ehk=TdcWPzr5xGP8xUOSOqZXauGOS1jHDMu7WnxPzkl7esw%3d
A body initially at rest travels a distance 100 m in 5 s with a constant acceleration. calculate
(i) Acceleration
(ii) Final velocity at the end of 5 s.
Answer:
(i)8m/s²(ii)40m/s
Explanation:
according to the formula
½at²=s.
then substituting the data
½a•5²=100
a=8m/s²
v=at=8•5=40m/s
Answer:
(I)
[tex]{ \bf{s = ut + \frac{1}{2} a {t}^{2} }} \\ 100 = (0 \times 5) + \frac{1}{2} \times a \times {5}^{2} \\ 200 = 25a \\ { \tt{acceleration = 8 \: m {s}^{ -2} }}[/tex]
(ii)
[tex]{ \bf{v = u + at}} \\ v = 0 + (8 \times 5) \\ { \tt{final \: velocity = 40 \: m {s}^{ - 1} }}[/tex]
Which one of the following is not an example of convection? An eagle soars on an updraft of wind. A person gets a suntan on a beach. An electric heater warms a room. Smoke rises above a fire. Spaghetti is cooked in water.
Answer: The statement that is not an example of convection is (A person gets a suntan on a beach).
Explanation:
There are different modes of heat energy transfer which includes:
--> conduction
--> Radiation and
--> Convection
CONVECTION is a process by which heat energy is transferred in a fluid or air by the actual movement of the heated molecules. The cooler portion of the air surrounding a warmer part exerts a buoyant force on it. As the warmer part of the air moves, it is replaced by cooler air that is subsequently warmed.
Convection in gases is very common and gas expands more than liquid when subjected to high temperature.
--> it is used in bringing about the circulation of fresh air in the room in a process known as ventilation.Here, cool air is constantly being replaced with denser air ( warm air).
-->An electric heater warms a room and Smoke rises above a fire are typical example of convection in gases.
-->Spaghetti is cooked in water: As the water close to the burner warms, it rises to the top and boils. At the same time, cooler water on top moves downward to replace the rising hot water.
--> also the eagle uses convection current to stay afloat in the sky without flapping its wings to conserve energy.
But the option (A person gets a suntan on a beach) is an example of heat transfer through radiation. This is because the sun emits it's rays from the sky down to earth without any material medium unlike others. Therefore, this option is the ODD one out.
State whether plastic is biodegradable or non-biodegradable ? Give reasons for your answer.
Answer:
non biodegradable
Explanation:
It is non biodegradable because plastic cannot dispose off easily ..
A planet of mass m moves around the Sun of mass M in an elliptical orbit. The maximum and minimum distance of the planet from the Sun are r1 and r2, respectively. Find the relation between the time period of the planet in terms of r1 and r2.
Answer:
the relation between the time period of the planet is
T = 2π √[( r1 + r2 )³ / 8GM ]
Explanation:
Given the data i the question;
mass of sun = M
minimum and maximum distance = r1 and r2 respectively
Now, using Kepler's third law,
" the square of period T of any planet is proportional to the cube of average distance "
T² ∝ R³
average distance a = ( r1 + r2 ) / 2
we know that
T² = 4π²a³ / GM
T² = 4π² [( ( r1 + r2 ) / 2 )³ / GM ]
T² = 4π² [( ( r1 + r2 )³ / 8 ) / GM ]
T² = 4π² [( r1 + r2 )³ / 8GM ]
T = √[ 4π² [( r1 + r2 )³ / 8GM ] ]
T = 2π √[( r1 + r2 )³ / 8GM ]
Therefore, the relation between the time period of the planet is
T = 2π √[( r1 + r2 )³ / 8GM ]
Una pelota se lanza verticalmente hacia arriba desde la azotea de un edificio con una velocidad inicial de 35 m/s. Si se detiene en el aire a 200 m del suelo, ¿Cuál es la altura del edificio?
a. 138,8 m
b. 51.2 m
c. 71,2 m
d. 45,0 m
Một mặt phẳng vô hạn tích điện đều, mật độ σ = 4.10-9 C/cm2 , đặt thẳng đứng trong không khí. Một quả cầu nhỏ có khối lượng 8 g, mang điện tích q = 10-8 C treo gần vào mặt phẳng, sao cho dây treo lúc đầu song song với mặt phẳng. Lấy g = 9,8m/s2 . Khi cân bằng, dây treo quả cầu hợp với mặt phẳng 1 góc bằng bao nhiêu?
Answer:
The angle is 16 degree.
Explanation:
A uniformly charged infinite plane, density σ = 4.10-9 C/cm2, is placed vertically in air. A small ball of mass 8 g, with charge q = 10-8 C, hangs close to the plane, so that the string is initially parallel to the plane. Take g = 9.8m/s2 . In equilibrium, by what angle does the string hanging from the ball make an angle with the plane?
Surface charge density, σ = 4 x 10^-5 C/m^2
charge, q = 10^-8 C
mass, m = 0.008 kg
The electric field due to the plate is
[tex]E= \frac{\sigma }{2\varepsilon 0}[/tex]
Let the angle make with the vertical is A and T is the tension in the string.
[tex]T sin A = q E....(1)\\\\T cos A = m g .... (2)\\\\Divie (1) by (2)\\\\tan A =\frac{q E}{m g}\\\\tan A = \frac{10^{-8}\times 4\times 10^{-5}}{2\times 8.85\times 10^{-12}\times 0.008\times9.8}\\\\tan A = 0.288\\\\A = 16 degree\\[/tex]
A pilot flies her route in two straight-line segments. The displacement vector A for the first segment has a magnitude of 243 km and a direction 30.0o north of east. The displacement vector for the second segment has a magnitude of 178 km and a direction due west. The resultant displacement vector is R = A + B and makes an angle ? with the direction due east. Using the component method, find (a) the magnitude of R and (b) the directional angle ?.
(a) R = km
(b) ? = degrees
Answer:
a) [tex]R=126Km[/tex]
b) [tex]\theta=74.6\textdegree[/tex]
Explanation:
From the question we are told that:
1st segment
243km at Angle=30
2nd segment
178km West
Resolving to the X axis
[tex]F_x=243cos30+178[/tex]
[tex]F_x=33.44Km[/tex]
Resolving to the Y axis
[tex]F_y=243sin30+178sin0[/tex]
[tex]R=\sqrt{F_y^2+F_x^2}[/tex]
[tex]F_y=121.5Km[/tex]
Therefore
Generally the equation for Directional Angle is mathematically given by
[tex]\theta=tan^{-1}\frac{F_y}{F_x}[/tex]
[tex]\theta=tan^{-1}\frac{121.5}{33.44}[/tex]
[tex]\theta=74.6\textdegree[/tex]
Generally the equation for Magnitude is mathematically given by
[tex]R=\sqrt{F_y^2+F_x^2}[/tex]
[tex]R=\sqrt{33.44^2+121.5^2}[/tex]
[tex]R=126Km[/tex]
Two charged particles attract each other with a force of magnitude F acting on each. If the charge of one is doubled and the distance separating the particles is also doubled, the force acting on each of the two particles has magnitude
(a) F/2,
(b) F/4,
(c) F,
(d) 2F,
(e) 4F,
(f) None of the above.
Answer:
F/2
Explanation:
In the first case, the two charges are Q1 and Q2 and the distance between them is r. K is the Coulomb's constant
Hence;
F= KQ1Q2/r^2 ------(1)
Where the charge on Q1 is doubled and the distance separating the charges is also doubled;
F= K2Q1 Q2/(2r)^2
F2= 2KQ1Q2/4r^2 ----(2)
F2= F/2
Comparing (1) and (2)
The magnitude of force acting on each of the two particles is;
F= F/2
a vehicle start moving at 15m/s. How long will it take to stop at a distance of 15m?
Speed= distance/time
Or time = distance/speed
According to your question
Speed=15m/s
and. Distance=1.2km. ,we must change kilometer in meter because given speed is in m/s
D= 1.2km = 1.2×1000m =1200meter
Time = distance/ speed
1200/15 =80second
Or. 1min and 20 sec will be your answer.
What is the effect on range and maximum height of a projectile as the launch height, launch speed, and launch angle are increased?
Answer:
The highest point in the trajectory occurs at the midpoint of the path. This highest point increases as the angle increases. At a 75° launch angle, the maximum height is approximately 76 meters. However, a further increase in launch angle beyond this 75° angle will increase the peak height even more.
I need help with this problem can anybody help me please , it’s physics 2 course
Answer:
ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|
Explanation:
The 2 capacitors in the middle are connected in parallel so simply add their capacitance together:
[tex]5.0\:\mu\text{F} + 8.0\:\mu\text{F} = 13.0\:\mu \text{F}[/tex]
Now we have 3 capacitors connected in series so their equivalent capacitance [tex]C_{eq}[/tex] is
[tex]\dfrac{1}{C_{eq}} = \dfrac{1}{10.0\:\mu \text{F}} + \dfrac{1}{13.0\:\mu \text{F}} + \dfrac{1}{9.0\:\ mu \text{F}} [/tex]
or
[tex]C_{eq} = 3.5\:\mu \text{F}[/tex]
If Katie swims from one end of the pool, to the other side, and then swims back to her original spot, her average velocity is half of her average speed when she swam to the other side.a) trueb) false
Answer:
false.
Explanation:
Ok, we define average velocity as the sum of the initial and final velocity divided by two.
Remember that the velocity is a vector, so it has a direction.
Then when she goes from the 1st end to the other, the velocity is positive
When she goes back, the velocity is negative
if both cases the magnitude of the velocity, the speed, is the same, then the average velocity is:
AV = (V + (-V))/2 = 0
While the average speed is the quotient between the total distance traveled (twice the length of the pool) and the time it took to travel it.
So we already can see that the average velocity will not be equal to half of the average speed.
The statement is false
A 1,760 W toaster, a 1,420 W electric frying pan, and an 85 W lamp are plugged into the same outlet in a 15 A, 120 V circuit. (The three devices are in parallel when plugged into the same socket.) (a) What current (in A) is drawn by each device
Answer:
Toaster = I = 14.67 A
Frying Pan = 11.83 A
Lamp = 0.71 A
Explanation:
The electric power is given as:
[tex]P = VI\\\\I = \frac{P}{V}[/tex]
where,
I = current
P = Power
V = Voltage = 120 V
FOR TOASTER:
P = 1760 W
Therefore,
[tex]I = \frac{1760\ W}{120\ V}[/tex]
I = 14.67 A
FOR FRYING PAN:
P = 1420 W
Therefore,
[tex]I = \frac{1420\ W}{120\ V}[/tex]
I = 11.83 A
FOR LAMP:
P = 85 W
Therefore,
[tex]I = \frac{85\ W}{120\ V}[/tex]
I = 0.71 A
Differences between angle of twist and angle of shear
Answer:
idek
Explanation:
Puck B has twice the mass of puck A. Starting from rest, both pucks are pulled the same distance across frictionless ice by strings with the same tension.a. Compare the final kinetic energies of pucks A and B. b. Compare the final speeds of pucks A and B.
Answer:
(a) 1 : 2
(b) same
Explanation:
Let the mass of puck A is m and the mass of puck B is 2 m.
initial speed for both the pucks is same as u and the distance is same for both is s.
let the tension is T for same.
The kinetic energy is given by
[tex]K = 0.5 mv^2[/tex]
(a) As the speed is same, so the kinetic energy depends on the mass.
So, kinetic energy of A : Kinetic energy of B = m : 2m = 1 : 2
(b) A the distance s same so the final velocities are also same.
(a) The kinetic energy of puck B is 2 times the kinetic energy of puck A.
(b) The final speed of both the puck A and B are same.
Let the mass of puck A is m and the mass of puck B is 2 m.
Initial speed for both the pucks is same as u and the distance is same for both is s.
Let the tension is T for same.
Then, the kinetic energy is given as,
[tex]KE = \dfrac{1}{2}mv^{2}[/tex]
(a)
As the speed is same, so the kinetic energy depends on the mass.
Then,
[tex]\dfrac{KE_{A}}{KE_{B}} = \dfrac{1/2 \times mv^{2}}{1/2 \times (2m)v^{2}}\\\\\\\dfrac{KE_{A}}{KE_{B}} =\dfrac{1}{2}[/tex]
So, kinetic energy of A : Kinetic energy of B = 1 : 2.
Thus, we can conclude that the kinetic energy of puck B is 2 times the kinetic energy of puck A.
(b)
The final speed for the puck is given as,
v = s/t
here, s is the distance covered.
Since, both pucks are pulled the same distance across frictionless ice. Then, the final speed of each puck is also same.
Thus, we can conclude that the final speed of both the puck A and B are same.
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A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?
Answer:
The correct answer is "[tex]4.49\times 10^{10} \ joules[/tex]".
Explanation:
According to the question,
The work will be:
⇒ [tex]Work=-\frac{kQq}{R}[/tex]
[tex]=-\frac{1}{4 \pi \varepsilon \times (18-30)\times 3\times 0.2}[/tex]
[tex]=-\frac{1}{4 \pi \varepsilon \times (-12)\times 3\times 0.2}[/tex]
[tex]=\frac{0.3978}{\varepsilon }[/tex]
[tex]=4.49\times 10^{10} \ joules[/tex]
Thus the above is the correct answer.
We have that the workdone is mathematically given as
W=4.49*10e10 J
From the question we are told
A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?WorkdoneGenerally the equation for the workdone is mathematically given as
W=-kQq/R
Therefore
0.3978/ε0 =-1/(4πε0*(18-30)*3*0.2
Hence
W=4.49*10e10 JFor more information on Charge visit
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why is unit of power is called derived unit?
Distance travelled by a body in unit time is called speed. it is a scalar quantity because it can be specified only by magnitude.
A charge Q exerts a 1.2 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of the force exerted on Q by q
Answer:
0.3 N
Explanation:
Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.
The magnitude of the force exerted on Q by q when the distance between them is doubled is 0.3 N
Coulomb's law equationF = Kq₁q₂ / r²
Where
F is the force of attraction K is the electrical constant q₁ and q₂ are two point charges r is the distance apart Data obtained from the question Initial distance apart (r₁) = rInitial force (F₁) = 1.2 NFinal distance apart (r₂) = 2rFinal force (F₂) =? How to determine the final forceFrom Coulomb's law,
F = Kq₁q₂ / r²
Cross multiply
Fr² = Kq₁q₂
Kq₁q₂ = constant
F₁r₁² = F₂r₂²
With the above formula, we can obtain the final force as follow:
F₁r₁² = F₂r₂²
1.2 × r² = F₂ × (2r)²
1.2r² = F₂ × 4r²
Divide both side by 4r²
F₂ = 1.2r² / 4r²
F₂ = 0.3 N
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What does Boyle's Law state about the relationship between the pressure and volume of an ideal gas at constant temperature?
a) The product of pressure and volume increases as pressure decreases.
b) The sum of pressure and volume is constant.
c) The sum of product and volume decreases as volume increases.
d) The product of pressure and volume is constant.
Answer:
Option (d).
Explanation:
According to the Boyle's law, for a given mass of a gas, the pressure of the gas is inversely proportional to the volume of the gas keeping the temperature of the gas is constant.
So,
Let the pressure is P, volume is V and T is the absolute temperature of the gas.
Pressure proportional to the reciprocal of the volume.
[tex]P \alpha \frac{1}{V}\\\\P V = constant[/tex]
The correct option is (d).
A random sample of 22 lunch orders at Noodles & Company showed a mean bill of $10.26
with a standard deviation of $5.21. Find the 99 percent confidence interval for the mean bill of
all lunch orders.
Answer:
(7.115 ; 13.405)
Explanation:
Given :
Sample size, n = 22
Mean bill, μ = 10.26
Standard deviation, s = 5.21
To obtain the 99% confidence interval for the mean bill of all orders ;
Mean ± margin of error
Margin of Error = Tcritical * s/√n
Tcritical at 99%, df = n-1, 22 - 1 = 21
Tcritical = 2.831
Margin of Error = 2.831 * (5.21/√22) = 3.145
Confidence interval = 10.26 ± 3.145
Lower boundary = 10.26 - 3.145 = 7.115
Upper boundary = 10.26 + 3.145 = 13.405
Confidence interval :
(7.115 ; 13.405)
Test your prediction through calculation for the situations of the clay bob and the bouncy ball. Assume each has a mass of 100 grams, and each has an initial velocity of 20 m/s straight at the door. Ignore the effects of gravity. Calculate the change in momentum of
Answer:
a) Δp = -2.0 kgm / s, b) Δp = -4 kg m / s
Explanation:
In this exercise the change in moment of a ball is asked in two different cases
a) clay ball, in this case the ball sticks to the door and we have an inelastic collision where the final velocity of the ball is zero
Δp = p_f - p₀
Δp = 0 - m v₀
Δp = - 0.100 20
Δp = -2.0 kgm / s
b) in this case we have a bouncing ball, this is an elastic collision, as the gate is fixed it can be considered an object of infinite mass, therefore the final speed of the ball has the same modulus of the initial velocity, but address would count
v_f = - v₀
Δp = p_f -p₀
Δp = m v_f - m v₀
Δp = m (v_f -v₀)
Δp = 0.100 (-20 - 20)
Δp = -4 kg m / s
A point charge is positioned in the center of a hollow metallic shell of radius R. During four experiments the value of the point charge and the total charge on shell were, respectively:
Complete question is;
A point charge is positioned in the center of a hollow metallic shell of radius R. During four experiments the value of point charge and charge of the shell were respectively:
+5q; 0
-6q; +2q
+2q; -3q
-4q; +12q
Rank the results of experiments according to the charge on the inner surface of the shell, most positive first:
a. 2, 3, 1, 4
b. 1, 2, 3, 4
c. 2, 4, 3, 1
d. 1, 3, 4, 2
Answer:
c. 2, 4, 3, 1
Explanation:
In this question, we can say that;
q_in = q_b
Where;
q_in is the charge on the inner surface of the shell
q_b is the point charge on the shell.
Thus q_in = -q_b was written because, as the shell is conducting, it means that the electric field would have a value of zero and thus the radius inside will be zero.
Thus;
- For +5q; 0:
q_in = -(+5q)
q_in = -5q
- For -6q; +2q :
q_in = - (-6q)
q_in = +6q
- For +2q; -3q :
q_in = -(+2q)
q_in = -2q
- For -4q; +12q:
q_in = -(-4q)
q_in = +4q
Ranking the most positive to the least positive ones, we have;
+6q, +4q, -2q, -5q
This corresponds to options;
2, 4, 3, 1
If 5.4 J of work is needed to stretch a spring from 15 cm to 21 cm and another 9 J is needed to stretch it from 21 cm to 27 cm, what is the natural length (in cm) of the spring
Answer:
the natural length of the spring is 9 cm
Explanation:
let the natural length of the spring = L
For each of the work done, we set up an integral equation;
[tex]5.4 = \int\limits^{21-l}_{15-l} {kx} \, dx \\\\5.4 = [\frac{1}{2}kx^2 ]^{21-l}_{15-l}\\\\5.4 = \frac{k}{2} [(21-l)^2 - (15-l)^2]\\\\k = \frac{2(5.4)}{(21-l)^2 - (15-l)^2} \ \ \ -----(1)[/tex]
The second equation of work done is set up as follows;
[tex]9 = \int\limits^{27-l}_{21-l} {kx} \, dx \\\\9 = [\frac{1}{2}kx^2 ]^{27-l}_{21-l}\\\\9 = \frac{k}{2} [(27-l)^2 - (21-l)^2] \\\\k = \frac{2(9)}{(27-l)^2 - (21-l)^2} \ \ \ -----(2)[/tex]
solve equation (1) and equation (2) together;
[tex]\frac{2(9)}{(27-l)^2 - (21-l)^2} = \frac{2(5.4)}{(21-l)^2 - (15-l)^2}\\\\\frac{2(9)}{2(5.4)} = \frac{(27-l)^2 - (21-l)^2}{(21-l)^2 - (15-l)^2}\\\\\frac{9}{5.4} = \frac{(729 - 54l+ l^2) - (441-42l+ l^2)}{(441-42l+ l^2) - (225 -30l+ l^2)} \\\\\frac{9}{5.4 } = \frac{288-12l}{216-12l} \\\\\frac{9}{5.4 } =\frac{12}{12} (\frac{24-l}{18 -l})\\\\\frac{9}{5.4 } = \frac{24-l}{18 -l}\\\\9(18-l) = 5.4(24-l)\\\\162-9l = 129.6-5.4l\\\\162-129.6 = 9l - 5.4 l\\\\32.4 = 3.6 l\\\\l = \frac{32.4}{3.6} \\\\[/tex]
[tex]l = 9 \ cm[/tex]
Therefore, the natural length of the spring is 9 cm
A person jumps out of an airplane above the surface of the Earth, and falls a distance h before opening their parachute. Once the prachute is open the person coasts to the ground a distance d at constant velocity.
a. The work done on the person by the Earth is:
b. The change in gravitational potential energy of the person + Earch system is:
Answer:
a) W_total = mg (2h + d) , b) E_total = - mg (h + d)
Explanation:
a) We must solve this problem in two parts, the first for the accelerated movement and the second for the movement with constant speed
Let's look for work for the part that is in free fall
y = y₀ + v₀ t - ½ g t²
when he jumps out of a plane his vertical speed is zero
y =y₀ - ½ g t²
dy = 0 - ½ g 2t dt
the work in this first part is
W₁ = ∫ F dy
W₁ = mg ∫ g t dt
W₁ = m g² t² / 2
the time it takes to travel the distance y₀-y = h is
y₀-y = ½ g t²
t =[tex]\sqrt{2h/g}[/tex]
we substitute
W₁ = m g² 2h / g
W₁ = m g 2h
now we look for the work for the part with constant speed
since the velocity is constant let's use the uniform motion ratio
W₂ = F d
W₂ = mg d
the total work is
W_total = W₁ + W₂
W_total = 2mgh + m gd
W_total = mg (2h + d)
b) The change in gravitational potential energy
U = mg Δy
in the part with accelerated movement
U₁ = mg h
in the part with uniform movement
U₂ = mg d
the total potential energy is
E_total = U₁ + U₂
E_total = - mg (h + d)
A wave moves in a rope with a certain wavelength. A second wave is made to move in the same rope with twice the wavelength of the first wave. The frequency of the second wave is _______________ the frequency of the first wave.
Answer:
The frequency of the second wave is half of the frequency of first one.
Explanation:
The wavelength of the second wave is double is the first wave.
As we know that the frequency is inversely proportional to the wavelength of the velocity is same.
velocity = frequency x wavelength
So, the ratio of frequency of second wave to the first wave is
[tex]\frac{f_2}{f_1} =\frac{\lambda _1}{\lambda _2}\\\\\frac{f_2}{f_1} =\frac{\lambda _1}{2\lambda _1}\\\\\frac{f_2}{f_1} =\frac{1}{2}\\\\[/tex]
The frequency of the second wave is half of the frequency of first one.